# 14. Calculate v, and the degree of inhibi- tion caused by a competitive inhibitor under the...

14. Calculate v, and the degree of inhibi- tion caused by a competitive inhibitor under the following conditions: (a) [S]-2x10" M and [I]-2x10-3 M, (b) [S]= 4x10-4 M and CI2x 10' M, and (c) [S]- 7.5x 10M and [I)-10° M, Assume that ..-2x10-3 M, K,=1·5 x 10" М, and Vmax = 270 nmoles × liter 1 x min-1.

Vi= VmaxS/Km+S[1+I/Ki]

Degree of inhibition= 1+(I/Ki)

Vmax= 270nmoles/litre/min

Km = 0.002M

Ki= 0.00015M

1) I= 0.002, S= 0.002

Vi= 270×0.002/0.002+0.002[1+0.002/0.00015]

=0.54/0.002+(0.002×14.33)

=0.54/[0.002+0.028]

=0.54/0.03

Vi=18 nmoles/litre/min

Degree of inhibition alpha= 1+(I/Ki)

= 1+(0.002/0.00015)

= 14.33

2) I= 0.00002, S= 0.0004

Vi= 270× 0.0004/0.002+0.0004[1+0.00002/0.00015]

= 0.108/0.002+(0.0004×1.13)

= 0.108/(0.002+0.00045)

= 0.108/0.0024

Vi= 45 nmoles/litre/min

Degree of inhibition alpha= 1+(0.00002/0.00015)

= 1.13

3) I= 0.00005, S= 0.0075

Vi= (270×0.0075)/0.002+0.0075[1+(0.00005/0.00015)]

= 2.025/0.002+0.0075(0.0002/0.00015)

= 2.025/0.002+0.0075×1.33

= 2.025/0.002+0.01

= 2.025/0.012

Vi= 168.75 nmoles/litre/min

Degree of inhibition alpha= 1+(0.00005/0.00015)

= 0.0002/0.00015

= 1.33

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